Math Pages Blog: Cats are the best....

Thursday, July 17, 2008

Cats are the best....

I usually don't see dreams, and I see dreams about mathematics only very rarely. However, this night I had a dream about mathematics, and my cat managed to turn it into the most bizarre dream I ever had... The first part of this post is the story itself, hope you will find it funny, and the second part is about power series.

In my dream I had a conversation about sums of infinite power series. I have no idea why I would see a dream about them, it is not like I was thinking much about this. Just when the conversation turned to a specific simple problem, my cat decieded that he wants my pillow. In other words he came and sat down on it (I guess I should be thankful that he didn't sit on my head...). Naturally, this waked my up and I attempted to remove him, while still half emerged in the dream. This effectively turned the conversation I had in my dream in a conversation with my cat. Eventually I manged to convince him to get off (not before he tried to bite me for this).
I have always claimed that while dogs are "mens best friend", cats are "mathematician best friend". Now I am sure in this. Somewhere deep inside them hides a mathematical mind... :)

Photo by Per Ola Wiberg..(PO...or Powi)

Now, to the problem I mentioned. It shall be named the dreaming cat problem (for no reason whatsoever). The problem is as follows: Suppose you have an infinite sum:

$\sum a_{n}x^n$

Suppose also that the sum converges for all x such that |x|<>R. R is a fixed positive number. If you will write instead of x, M-x (M is a fixed number) what will happen, if anything? The answer - absolutely nothing interesting. The only thing that will change is that now the sum will converge for all x such that |M-x| $a_{n}$ to $\frac{1}{a_{n}}$ ? This is more interesting, we will now get that the series converges for |x|< $R_{2}$ =< $\frac{1}{R}$ .

This is result is not obvious, so it is time to proof all what I written. Firstly, lets proof that there is indeed such R. To do this we will use the root exam for infinite sums. This exam says that any given infinite sum converges if $limsup|^{n}\sqrt{a_{n}}|<1$ (The limit is when n approaches infinity) and it doesn't converge if the limit is greater than one. If we will take this upper limit from the series we have we will get $limsup|^{n}\sqrt{a_{n}}|<\frac{1}{x}$ .

If we will suppose that the right side is equal to $\frac{1}{R}$ we will get that the sum converges for all |x|R.

To proof the second part, we will first need to proof a different statement. Given two different infinite sums $a_{n}x^n$ and $b_{n}x^n$ which converge for $R_{1}$ and $R_{2}$ , for what x the sum $(a_{n}b_{n}x^n$ ) converges? Lets use the above method:

$\frac{1}{R}=limsup|^n\sqrt{a_{n}b_{n}}|=<limsup|^n\sqrt{a_{n}}|*limsup|^n\sqrt{b_{n}}|=\frac{1}{R_{1}}\frac{1}{R_{1}}$

So we get that $R_{1}R_{2}=<R$ .
We can now use this formula to solve the problem. Lets suppose that $b_{n}=\frac{1}{a_{n}}$ for all n. Then $a_{n}b_{n}=1$ . From this we will get that R in the above formula is 1. Now if we will divide by $R_{1}$ we will get that $R_{2}<=\frac{1}{R_{1}}$ But $R_{1}$ is exactly R from the statement I wanted to proof, so we get that indeed this new series will converge for all |x|< $R_{2}$ =< $\frac{1}{R}$ .